Problem: What is the value of $\dfrac{d}{dx}\sec(x)$ at $x=\pi$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-1$ (Choice B) B $1$ (Choice C) C $\dfrac{1}{2}$ (Choice D) D $0$
Explanation: Let's first find $\dfrac{d}{dx}\sec(x)$. Then, we can evaluate it at $x=\pi$. Recall that the derivative of $\sec(x)$ is $\dfrac{\sin(x)}{\cos^2(x)}$, or $\sec(x)\tan(x)$. Put another way, $\dfrac{d}{dx}[\sec(x)]=\dfrac{\sin(x)}{\cos^2(x)}=\sec(x)\tan(x)$. [Is there a way to know this without memorizing?] Now let's plug in $x={\pi}$ : $\begin{aligned} &\phantom{=}\dfrac{\sin\left({\pi}\right)}{\cos^2\left({\pi}\right)} \\\\ &=\dfrac{0}{(-1)^2} \\\\ &=0 \end{aligned}$ In conclusion, the value of $\dfrac{d}{dx}\sec(x)$ at $x=\pi$ is $0$.